Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $y = \dfrac{t^2 + 7t}{t^2 + 3t - 28} \div \dfrac{2t^2 + 10t}{t^2 - 7t + 12} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{t^2 + 7t}{t^2 + 3t - 28} \times \dfrac{t^2 - 7t + 12}{2t^2 + 10t} $ First factor out any common factors. $y = \dfrac{t(t + 7)}{t^2 + 3t - 28} \times \dfrac{t^2 - 7t + 12}{2t(t + 5)} $ Then factor the quadratic expressions. $y = \dfrac {t(t + 7)} {(t - 4)(t + 7)} \times \dfrac {(t - 4)(t - 3)} {2t(t + 5)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {t(t + 7) \times (t - 4)(t - 3) } { (t - 4)(t + 7) \times 2t(t + 5)} $ $y = \dfrac {t(t - 4)(t - 3)(t + 7)} {2t(t - 4)(t + 7)(t + 5)} $ Notice that $(t - 4)$ and $(t + 7)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {t\cancel{(t - 4)}(t - 3)(t + 7)} {2t\cancel{(t - 4)}(t + 7)(t + 5)} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $y = \dfrac {t\cancel{(t - 4)}(t - 3)\cancel{(t + 7)}} {2t\cancel{(t - 4)}\cancel{(t + 7)}(t + 5)} $ We are dividing by $t + 7$ , so $t + 7 \neq 0$ Therefore, $t \neq -7$ $y = \dfrac {t(t - 3)} {2t(t + 5)} $ $ y = \dfrac{t - 3}{2(t + 5)}; t \neq 4; t \neq -7 $